2
$\begingroup$

Some functions are not represented by their power series even when they are continuous and have all the necessary derivatives. What's the best characterization of these functions? Explanations at any level are welcome.

$\endgroup$
3
  • 2
    $\begingroup$ mathoverflow.net/questions/72/… $\endgroup$ Nov 2, 2009 at 23:17
  • 3
    $\begingroup$ I suppose with the right topology, "most" C^infty functions are nowhere analytic in the sense of category. So maybe the right question is to characterize the analytic functions among the C^infty functions. $\endgroup$ Nov 2, 2009 at 23:42
  • $\begingroup$ @Qiaochu: I don't want a single example. I want to know what the best characterization of these functions is. $\endgroup$
    – Kim Greene
    Nov 3, 2009 at 0:04

2 Answers 2

6
$\begingroup$

A smooth function is characterized as being analytic if its derivatives on any closed interval have a certain growth rate. See "Alternate Characterizations" in the Wikipedia article on analytic functions.

$\endgroup$
0
$\begingroup$

Functions are represented by their power series if and only if they are analytic, i.e. complex differentiable.

Power series are easier to understand as functions of complex variables. For example, there's no apparent reason why the power series of 1/(1 + x^2) centered at 0 should have radius of convergence 1. It's infinitely differentiable everywhere. But as a function of a complex variable, it has a singularity at x = i, and that's why the radius of convergence is 1.

$\endgroup$
1
  • 2
    $\begingroup$ While valuable, I think as phrased this perspective (particularly the first sentence) can be a bit confusing, since it's not clear how to define "complex differentiable" for a function which a priori is defined only for real inputs. You have to talk about the existence of a complex analytic continuation to give this statement meaning, and I think already things are more conceptually loaded than if one just worked with the definition of real analytic. $\endgroup$ Nov 3, 2009 at 5:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.